Results 1 to 6 of 6

Math Help - Recurrence Relation

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    4

    Recurrence Relation

    Here is a question that I have been stuck on.

    A sequence is defined by the following recursive equation.
    Xn+1=√(2-xn) X0=0

    NOTE: the n+1 is supposed to be subscript and so is the last n as well as the 0 after the x.

    (a) Calculate the values of xn for n = 1, 2, 3 and 4. Report an exact answer and a decimal
    approximation.
    (b) Use mathematical induction to show that the even-numbered terms are increasing, that is,
    that: x0 < x2 < x4 < x6 etc.
    (c) Show that the even-numbered terms are all less than
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,854
    Thanks
    321
    Awards
    1
    Quote Originally Posted by turtlejacks View Post
    Here is a question that I have been stuck on.

    A sequence is defined by the following recursive equation.
    Xn+1=√(2-xn) X0=0

    NOTE: the n+1 is supposed to be subscript and so is the last n as well as the 0 after the x.

    (a) Calculate the values of xn for n = 1, 2, 3 and 4. Report an exact answer and a decimal
    approximation.
    (b) Use mathematical induction to show that the even-numbered terms are increasing, that is,
    that: x0 < x2 < x4 < x6 etc.
    (c) Show that the even-numbered terms are all less than
    How far have you gotten? And your last question got cut off.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    4
    oh its supposed to say less then 1. And I solved part a.
    Im confused how to do part b
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,687
    Thanks
    617
    Hello, turtlejacks!

    \text{A sequence is de{f}ined by the following recursive equation:}

    . . . . x_{n+1}\:=\:\sqrt{2-x_n},\;\;x_0\,=\,0


    \text{(a)  Calculate the values of }x_n\text{ for }n = 1, 2, 3,4.
    . . . . \text{Report an exact answer and a decimal <br />
approximation.}

    \begin{array}{ccc}x_n & \text{Exact} & \text{Approx.} \\ \hline \\[-4mm]<br />
x_0 & 0 & 0 \\ \\[-3mm]<br />
x_1 & \sqrt{2} & 1.414213 \\ \\[-3mm]<br />
x_2 & \sqrt{2\!-\!\sqrt{2}} & 0.761367 \\ \\[-3mm]<br />
x_3 & \sqrt{2\!-\!\sqrt{2\!-\!\sqrt{2}} & 1.111140 \\ \\[-3mm]<br />
x_4 & \sqrt{2\!-\!\sqrt{2\!-\!\sqrt{2\!-\!\sqrt{2}}} & 0.942793\end{array}




    \text{(b) Use mathematical induction to show that the even-numbered}
    . . . . \text{terms are increasing; that is, that: }\:x_0 < x_2 < x_4 < x_6\;\hdots

    \text{(c)  Show that the even-numbered terms are all less than 1.}

    I have not solved these parts yet, but I have some observations.

    We have: . x \:=\:\sqrt{2 + \sqrt{2 + \sqrt{2 + \hdots }}}

    That is: . x \:=\:\sqrt{2 + x} \quad\Rightarrow\quad x^2 \:=\:2+x \quad\Rightarrow\quad x^2 + x - 2 \:=\:0

    . . . (x - 1)(x + 2) \:=\:0 \quad\Rightarrow\quad x \:=\:1,-2

    Since \,x is obviously positive: . \sqrt{2 + \sqrt{2+ \sqrt{2 + \hdots}}}  \:=\: 1



    Extend and examine the table in (a).

    We find that the odd-numbered terms are approaching 1 from above,
    . . and that the even-numbered terms are approaching 1 from below.


    Does that give anyone a jump-start?

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    For the induction part.
    Suppose that we know that a_{2K}<a_{2K+2}.

    Which implies that -a_{2K}>-a_{2K+2}

    By adding 2 and taking the square root we get:
    \sqrt{2-a_{2K}}>\sqrt{2-a_{2K+2}}

    Or a_{2K+1}>a_{2K+3}

    Now consider a_{2K+2}=\sqrt{2-a_{2K+1}} <\sqrt{2-a_{2K+3}}=a_{2K+4}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,854
    Thanks
    321
    Awards
    1
    Thank you, Plato. That one was bugging me.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Recurrence relation
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: October 15th 2011, 11:27 PM
  2. recurrence relation
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: October 18th 2010, 02:15 AM
  3. Recurrence Relation HELP
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: May 3rd 2009, 01:18 PM
  4. recurrence relation
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 15th 2009, 06:20 PM
  5. Recurrence relation
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: November 16th 2008, 08:02 AM

Search Tags


/mathhelpforum @mathhelpforum